1-find-ln-x-x-1-dx-2-calculate-0-1-ln-x-x-1-dx-

Question Number 85162 by mathmax by abdo last updated on 19/Mar/20

1) f i n d \int l n (\sqrt{x} + \sqrt{x + 1}) d x

2) c a l c u l a t e \int_{0}^{1} l n (\sqrt{x} + \sqrt{x + 1}) d x

Commented by mathmax by abdo last updated on 01/Apr/20

1) l e t f (t) = \int l n (t + \sqrt{x} + \sqrt{x + 1}) d x

w e h a v e f^{^{'}} (t) \int \frac{1}{t + \sqrt{x} + \sqrt{x + 1}} d x =_{\sqrt{x} = u} \int \frac{2 u d u}{t + u + \sqrt{u^{2} + 1}}

=_{u = s h (α)} 2 \int \frac{s h (α) c h (α) d α}{t + s h (α) + c h (α)} = \int \frac{s h (2 α)}{t + s h α + c h α} d α

= \int \frac{\frac{e^{2 α} - e^{- α}}{2}}{t + e^{α}} d α =_{e^{α} = z} \frac{1}{2} \int \frac{z^{2} - z^{- 2}}{t + z} \frac{d z}{z}

= \frac{1}{2} \int \frac{z^{2} - z^{- 2}}{z (t + z)} d z = \frac{1}{2} \int \frac{z^{4} - 1}{z^{3} (t + z)} d z l e t d e c o m p o s e

F (z) = \frac{z^{4} - 1}{z^{3} (z + t)} \Rightarrow F (z) = \frac{z^{4} - 1}{z^{4} + {t z}^{3}} = \frac{z^{4} + {t z}^{3} - {t z}^{3} - 1}{z^{4} + {t z}^{3}} = 1 - \frac{{t z}^{3} + 1}{z^{4} + {t z}^{3}}

w (z) = \frac{{t z}^{3} + 1}{z^{3} (z + t)} = \frac{a}{z} + \frac{b}{z^{2}} + \frac{c}{z^{3}} + \frac{d}{z + t}

c = \frac{1}{t}, d = \frac{1 - t^{4}}{- t^{3}} = \frac{t^{4} - 1}{t^{3}} \Rightarrow w (z) = \frac{a}{z} + \frac{b}{z^{2}} + \frac{1}{{t z}^{3}} + \frac{t^{4} - 1}{t^{3} (z + t)}

{l i m}_{z \to + \infty} z w (z) = t = a + d \Rightarrow a = t - d = t - \frac{t^{4} - 1}{t^{3}} = \frac{1}{t^{3}} \Rightarrow

w (z) = \frac{1}{t^{3} z} + \frac{b}{z^{2}} + \frac{1}{{t z}^{3}} + \frac{t^{4} - 1}{t^{3} (z + t)}

w (1) = 1 = \frac{1}{t^{3}} + b + \frac{1}{t} + \frac{t^{4} - 1}{t + 1} = \frac{1}{t^{3}} + \frac{1}{t} + \frac{t^{4} - 1}{t + 1}

= \frac{t + t^{3}}{t^{4}} + \frac{t^{4} - 1}{t + 1} = \frac{(t + t^{3}) (t + 1) + t^{8} - t^{4}}{t^{4} (t + 1)}

= \frac{t^{2} + t + t^{4} + t^{3} + t^{8} - t^{4}}{t^{4} (t + 1)} = \frac{t^{8} + t^{3} + t^{2} + t}{t^{4} (t + 1)} = \frac{t^{7} + t^{2} + t + 1}{t^{3} (t + 1)} \Rightarrow

b = 1 - \frac{t^{7} + t^{2} + t + 1}{t^{3} (t + 1)} w e h a v e F (z) = 1 - \frac{a}{z} - \frac{b}{z^{2}} - \frac{c}{z^{3}} - \frac{d}{z + t} \Rightarrow

\int F (z) d z = z - a l n ∣ z ∣ + \frac{b}{z} + \frac{c}{2 z^{2}} - d l n ∣ z + t ∣ + C \dots b e c o n t i n u e d \dots